Let $R$ be an $(n+2)$-ary relation symbol and let $\phi(v_1, \cdots v_n)$ be the formula $\exists x \forall y R(v_1, \cdots v_n, x,y)$. Consider a totally ordered set $(I, \leq)$ and $\mathcal{M}_i$ stuctures such that $\mathcal{M}_i$ is a substructure of $\mathcal{M}_j$ whenever $i \leq j$, and let $\mathcal{M}$ be the union of the $\mathcal{M}_i$ structures.
I am trying to show that there exists some $i$ such that, for all $b_1, \cdots b_n \in M_i$ satisfying $\mathcal{M} \models \phi[b_1 \cdots b_n]$, we have $\mathcal{M}_i \models \phi[b_1 \cdots b_n]$.
What I tried was as follows:
(1) Saying $\mathcal{M} \models \phi[b_1 \cdots b_n]$ is the same as saying there exists a $c \in M$ such that for all $d \in M$ we have $R^{\mathcal{M}}(b_1 \cdots b_n,c,d)$ is true.
(2) Saying $\mathcal{M}_i \models \phi[b_1 \cdots b_n]$ is the same as saying there exists a $c \in M_i$ such that for all $d \in M_i$ we have $R^{\mathcal{M}_i}(b_1 \cdots b_n,c,d)$ is true.
So it seems to me that we just need to make sure that the $c \in M$ of (1) can actually be chosen from $M_i$. Now $M= \bigcup_{j \in I} M_j$, so we can pick $c \in M_j$ for some $j$. But we choose $i$ before we can choose $c$, so it doesn't seem that we can just force $c \in M_i$.
This is not true. What is true is that for all $b_1,\dots,b_n\in M$ such that $M\models \varphi(b_1,\dots,b_n)$, there exists some $i$ such that $M_i\models \varphi(b_1,\dots,b_n)$. But the $i$ may have to depend on the choice of $b_1,\dots,b_n$.
For example, set $n = 1$, so $R$ has arity $3$. Our ordered set will be $(\omega,\leq)$. For each $i\in \omega$, let $M_i = \{0,\dots,i\}$. Define $R^{M_i} = \{(a,b,c)\mid a<b\}$. (Note that $c$ is irrelevant here, so $R$ is really a binary relation in disguise, and $\varphi(z)$ is equivalent to $\exists x\, (z<x)$ in all the structures $M_i$.) Then $M_i$ is a substructure of $M_j$ when $i\leq j$. Let $M = \bigcup_{i\in \omega} M_i$.
Now fix $j\in \omega$. We have $M\models \varphi(j)$, since $M\models R(j,j+1,k)$ for all $k$, but $M_j\not\models \varphi(j)$, since $j$ is maximal in the standard order on $M_j$.